If the solubility product of a sparingly soluble salt Li3PO4 is 2.7× – InterviewSolution

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1.	If the solubility product of a sparingly soluble salt Li3PO4 is 2.7×10−9. The concentration (mol L−1) of Li+ ions in the saturated solution of the salt is:Given : (100)14=3.16
Answer» If the solubility product of a sparingly soluble salt $L i_{3} P O_{4}$ is $2.7 \times 10^{- 9}$ . The concentration $(m o l L^{- 1})$ of $L i^{+}$ ions in the saturated solution of the salt is: Given : $(100)$ $^{\frac{1}{4}} = 3.16$

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